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What is the expected LCM of 21 randomly selected positive integers under 10000000?

How would someone even approach this problem?

EDIT: The positive integers are chosen with replacement.

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    $\begingroup$ I would approach it by seeing what happens with smaller numbers, in particular, what's the expected LCM for two random integers (less than whatever). Also, are you selecting random integers with or without replacement? I.e., can you select the same number more than once? $\endgroup$ – Barry Cipra Aug 26 '15 at 17:53
  • $\begingroup$ Thanks, I didn't think of that. By the way, you can select the same number more than once. $\endgroup$ – Ant Aug 26 '15 at 18:01
  • $\begingroup$ If $P(n$) is the chance that 2 random numbers less than $ n$ (with replacement) are co-prime then $ \lim_{n \to \infty} P(n)=6 \pi^{-2}$. $\endgroup$ – DanielWainfleet Aug 26 '15 at 18:11
  • $\begingroup$ What kind of answer do you want? Exact number? Algorithm to determine it? Asymptotic size as 10000000 goes to infinity? $\endgroup$ – Greg Martin Aug 26 '15 at 18:49
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Let $1 \leq a_1, \dots, a_n \leq N$ be number chosen uniformly at random. What is their least common multiple?

What is the highest power of $2$ dividing any of the $a_1, \dots, a_n$?

  • All the numbers are odd with probability $(1 - \frac{1}{2})^n$
  • All the numbers are odd or even (but not divisible by 4) with probability $(1 - \frac{1}{4})^n$
  • All the numbers are odd or even (but not divisible by 8) with probability $(1 - \frac{1}{8})^n$
  • ...

So the expected power of $2$ dividing all these numbers is (this number might simplify):

$$ \mathbb{E}_2 = \sum_{k \geq 0} \Big(2^k - 2^{k-1} \Big) \left[1- \left( 1- \tfrac{1}{2^k} \right)^n \right]$$

A similar story for $\mathbb{E}_3, \mathbb{E}_5, \mathbb{E}_7$ etc; multiply your answers

$$ \mathbb{E} = \prod_{p } \mathbb{E}_p$$


See also: Expectation of the maximum of i.i.d. geometric random variables a quick look confirms there is no closed-form answer.

In your case $n = 21$ and $N = 1,000,000$ so we can hope for an estimate.


One way to state the prime number theorem is that $\mathrm{lcm}(1,2,\dots, n) = e^n$ So the least common multiple is growing exponentially fast in $n$. In your case, there are $n = 21$ numbers ranging from $1$ to $N = 10^6$. Perhaps

$$ \mathbb{E} \approx e^{n} \times \left(\frac{N}{n}\right)^n \approx \frac{N^n}{n!} = \frac{10^{21}}{ 21!} \approx \frac{10^{21}}{ 5 \times 10^{19}} = 40 $$

Still doesn't seem quite right. See Granville Prime Number Races)

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  • $\begingroup$ $N^n$ is $(10^6)^21$, not $10^21$. This would lead to $2⋅10^{104}$. $\endgroup$ – Frédéric Grosshans Aug 2 '17 at 16:30
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Whether you chose the integer with replacement or not is unlikely to make a difference: the probability to have 21 different integers in your set of 21 is $$\exp\left(\sum_{i=0}^{20}\log\left(1-\frac i{10^7}\right) \right)\simeq 1 - \frac{20⋅21}{2⋅10^7}=1-2.1⋅10^{-5}$$.

A paper by Cilleruelo, Rué, Šarka and Zumalacárregui (arXiv:1112.3013 , J. Num. Th. 144 92) specifically addresses similar questions. They compute $\DeclareMathOperator\lcm{lcm}$ $$ψ(A)=\log\lcm\{a:a∈A\} $$ for random subsets $A⊆\{1…n\}$ of size $|A|=k(n)$ when $n→∞$. When $k(n)→∞$, their theorem 1.2 says that, almost surely $$ψ(A)=k\frac{\log\frac nk}{1-\frac kn}\left(1+O\left(e^{-C\sqrt{\log k}}\right)\right), $$ for a constant $C$.

In your case, you have $n=10^7$, $k=21$. Since $k$ is not really large, you should look into the paper to see how this is proven, and to which extent it applies to your case. Boldly assuming $k=21→∞$ (I know !), would led to

$$ψ(A)=21\frac{19.16}{1-2.1⋅10^{-6}}\left(1+O\left(e^{-1.74C}\right)\right) ∼ 400 \text{, if $C$ is big enough.} $$

In other words, we would have $\lcm(A)∼e^{400}∼5⋅10^{173}$ if my rough unjustified approximations are correct.

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